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Differential of exponential operator

  1. Apr 16, 2014 #1
    If [itex]\hat{U}(r) = e^{\hat{A}(r)}[/itex], can we say [itex]\frac{d\hat{U}}{dr} = \frac{d\hat{A}}{dr}e^{\hat{A}(r)}[/itex]?
     
  2. jcsd
  3. Apr 16, 2014 #2
  4. Apr 16, 2014 #3

    micromass

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    For matrices, this is true provided that ##A## is differentiable.

    The general (infinite-dimensional) case is a bit more annoying, especially if you let ##A(r)## be unbounded. So since I think you're only really interested in the case ##A(r) = irA## with ##A## self-adjoint (which is actually a fairly general case: http://en.wikipedia.org/wiki/Stone's_theorem_on_one-parameter_unitary_groups ), then you can indeed say that

    [tex]\frac{d U}{dr} = iA[/tex]

    in the sense that for each ##\psi## in the domain of ##A## holds that

    [tex]\lim_{r\rightarrow 0} \left\|\frac{U(r)\psi - \psi}{r} - iA\psi\right\| = 0[/tex]

    Books like Reed & Simon, and Hall's QM for mathematicians give proofs. A naive proof of expanding the exponential in a power series would probably work in the case that ##A## is a bounded operator, but likely fails in the unbounded case. You need the spectral theorem there (you already need it just to define the exponential).
     
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